问题补充:
求解常微分方程 y+2xy+(x^2)y\=0 坐等…
答案:
y+2xy+(x^2)y\=0
设x=e^t, t=lnx
y(x)=y(t)/x . xy(x)=y(t)
y\(x)=(y\(t)-y(t))/x^2 . x^2y\(x)=y\(t)-y(t)
y\(t)-y(t)+2y(t)+y=0 y\(t)+y(t)+y=0
解得:y=e^(-t/2)(C1cos(t√3/2)+C2sin(t√3/2))
=x^(-1/2)(C1cos(√3lnx/2)+C2sin(√3lnx/2))